a crane lifts a 75 kg mass at a height of 8m . calculate the gravitational potential energy gained by the mass (g= 10N)

A child throws a ball perfectly horizontally at the same time a dirt clod falls off it atthe same height above the ground. The c

Oksana_A [137]

Answer: A.). Be equal to the time the ball hits the group

Explanation:

6 0

3 years ago

How do whales bats and dolphins use echolocation?

g100num [7]

C is your answer ithink

3 0

3 years ago

Read 2 more answers

A camera lens used for taking close-up photographs has a focal length of 22.0 mm. The farthest it can be placed from the film is

Arada [10]

Answer:

p = 6.64 cm

Explanation:

For this exercise we use the equation of the constructor

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively

They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image

$\frac{1}{p} = \frac{1}{f} - \frac{1}{q}$

1 / p = 1 / 2.2 - 1/3.29

1 / p = 0.15059

p = 6.64 cm

therefore the farthest distance from the object is 6.64 c

3 0

3 years ago

A car that weighs 1.0 x 10^4 N is initially moving at a speed of 38 km/h when the brakes are applied and the car is brought to a

hram777 [196]

Answer:

Part a) Force on car = 2833.84 Newtons

Part b) Time to stop the car = 3.8 seconds

Part c) Factor for stopping distance is 4.

Part d) Factor for stopping time is 1.

Explanation:

The deceleration produced when the car is brought to rest in 20 meters can be found by third equation of kinematics as

$v^2=u^2+2as$

where

v = final speed of the car ( = 0 in our case since the car stops)

u = initial speed of the car = 38 km/hr = $\frac{38\times 1000}{3600}=10.56m/s$

a = deceleration produces

s = distance in which the car stops

Applying the given values we get

$0^2=10.56^2+2\times a\times 20\\\\a=\frac{0-10.56^2}{2\times 20}\\\\\therefore a=-2.78m/s^2$

Now the force can be obtained using newton's second law as

$Force=\frac{Weight}{g}\times a$

Applying values we get

$Force=\frac{1.0\times 10^4}{9.81}\times -2.78\\\\\therefore F=-2833.84Newtons$

The negative direction indicates that the force is opposite to the motion of the object.

Part b)

The time required to stop the car can be found using the first equation of kinematics as

$v=u+at$ with symbols having the same meanings

Applying values we get

$0=10.56-2.78\times t\\\\\therefore t=\frac{10.56}{2.78}=3.8seconds$

Part c)

From the developed relation of stopping distance we can see that the for same force( Same acceleration) the stopping distance is proportional to the square of the initial speed thus doubling the initial speed increases the stopping distance 4 times.

Part d)

From the relation of stopping time and the initial speed we can see that the stopping distance is proportional initial speed thus if we double the initial speed the stopping time also doubles.

8 0

3 years ago

Why will interference occur

den301095 [7]

Two waves interfere when they run into each other.

The barrier reflects waves that run straight into it. It acts as a wave source and sends wave pulses back up the page towards the incoming waves.

Imagine a loose string tied to a wall. Someone sends two consecutive pulses along the string towards the wall. The first pulse gets reflected right away. It will travel backward towards the person holding the string. Along its way, it will run into the second pulse. The two pulses will interfere. The wall will make the reflected pulse out of phase with the second one. They will end up creating a destructive interference.

So is the case with the water waves running into the barrier. The barrier will send incoming waves back toward where they came from. Reflected waves interfere with incoming ones.

7 0

3 years ago